The value of ' $a$ ' for which one root of the quadratic equation $\left(a^2-5 a+3\right) x^3+(3 a-1) x+2=0$ is twice as large as the other is

Question

Aicharya · Accepted Answer

$3 \alpha=\frac{1-3 a}{a^2-5 a+3} \& 2 \alpha^2=\frac{2}{a^2-5 a+3}$ $2\left[\frac{1}{9} \frac{(1-3 a)^2}{\left(a^2-5 a+3\right)^2}\right]=\frac{2}{a^2-5 a+3}$ $\frac{(1-3 a)^2}{\left(a^2-5 a+3\right)}=9$ or $9 a^2-6 a+1$ $=9 a^2-45 a+27$ or $39 a=26$ or $a=\frac{2}{3}$

The value of ' $a$ ' for which one root of the quadratic equation $\left(a^2-5 a+3\right) x^3+(3 a-1) x+2=0$…

Solution